NavList:

   

Reply

On 29 May 2001, at 15:13, Noyce, Bill wrote:
> > > > 4)  What is the Compass Course(CC)/Course-to-Steer for Aruba Gap
> > > >     from the DR position?
> > > > --  ------------------------------------------------------------
> > >
> > > Using Law of Sines, I find I need to adjust my course North
> > > about 1.5 degrees.
> > >   CC = 283.4 + 1.5 + 9 = 293.9d
> >
> > Mmmm..Barely OK. Try a current vector diagram.
>
> I drew a diagram to get the orientation of the pieces right, but I
> have a hard time measuring the resulting course adjustment, because
> it's so small.  Thus the attempt at a 'digital' method.
You can make the vector diagram any convenient size.
>
> TC=283.4, Set=265.0, angle between = 18.4 degrees
>
>   sin 18.4 / speed = sin adj / drift
>
>   sin 18.4 / 8.5 = sin adj / 0.7
>
> I did this with log tables before; doing it on the computer now
> says adj = 1.490 degrees.  The diagram indicates it needs to be
> added to the course.  Is there something inherantly wrong with
> this approach, or have I made a slip somewhere?
Not being mathematically oriented I can't say, but in Shufeldts's The
Calculator Afloat, he gives the following formulas:
  Finding correction angle and speed of advance
        Value of correction angle
                sin B = a x sin A /c
    Calculate speed of advance
                c = a x sin [d -(A + B)] /sin A
  Current Sailing when when track and speed of advance are specified
        Value of correction angle
                tan B = b x sin A / a - b x cos A
        Value of speed to steam
                a = b x sin A /sin B
> > >
> > > Mid-latitude using 13d 42'
> > > dLat = 15' N  dLon = 183d W  dep = 177.8 W
> > > TC = 274.6d  dist = 178.3 nm
> >
> > OK. Be aware that above 500 miles at 90d/270d Mid-Latitude can give an
> error.
> > Above 1200 mils it gets unreliable.
>
> I saw a discussion initiated by Sam Chan about Mercator sailing
> near 90 or 270, where the problem arises from rounding the course
> before using dLat/cos(C) to compute the total distance.  Is this
> a similar problem?  In Mid-Latitude sailing, where we already
> have dLat and departure, the total distance can be computed either
> as dLat/cos(C) or as departure/sin(C), so we don't have to divide
> by a tiny rounded number.
>
> Or are you talking about the error in distance that arises from
> assuming a spherical earth?  What method would be
> better? I assumed we weren't planning to sail a great-circle
> course here.
As I understand it it's caused by the formulas and the how they are derived.
Mid-Latitude gets error prone when you cross the sperical quadrants. Try
calculating using Mid-Latitude from North America to Australia and one N.AS.
to South America of about equal distance.
GC course comes after the Panama Canal transit ;^)
Dan Hogan WA6PBY
dhhogan@XXX.XXX
Nav-L Web Page: http://www.wa6pby.com
Catalina 27 "GACHA"

   

Reply